Show that every subgroup of the quaternion group is normal and find the isomorphism type of the corresponding quotient ?
I know that $Q_8$ has a subgroup $\langle i\rangle=\{1,i,-1,-i\}$, $\langle j\rangle=\{1,j,-1,-j\}$, $\langle k\rangle=\{1,k,-1,-k\}$, $\langle -1\rangle=\{1,-1\}$. So basically, I have to prove that everyone of these subgroups are a normal subgroup the isomorphism type of the corresponding quotient. Would anyone has an idea how I can get started. I want keep in mind that we have not gone over Lagrange's theorem yet and has not proved that every subgroup of index $2$ is normal.
Best Answer
It's easy to show that $\langle -1\rangle$ is normal, because the elements $1$ and $-1$ commute with each element.
Consider $\langle i\rangle$; you just need to prove $xix^{-1}\in\langle i\rangle$, for every $x\in Q_8$, because you know $x1x^{-1}=1\in \langle i\rangle$ and $x(-1)x^{-1}=-1\in\langle i\rangle$; also $x(-i)x^{-1}=-xix^{-1}$, so $x(-i)x^{-1}\in\langle i\rangle$ as soon as $xix^{-1}\in\langle i\rangle$.
The statement is obviously true for $x=1,-1,i,-i$. Try with $j$: $$ jij^{-1}=ji(-j)=-j(ij)=-jk=-i $$ Similarly, $$ kik^{-1}=ki(-k)=-(ki)k=-jk=-i $$
Apply the same for $\langle j\rangle$ and $\langle k\rangle$.
On the other hand, you don't need to know special theorems for deducing that a subgroup $N$ of index $2$ in a group $G$ is normal. Indeed, the left cosets are $N$ and $gN=G\setminus N$ (where $g\notin N$), whereas the right cosets are $N$ and $Ng=G\setminus N$. So, for any $x\in N$, $xN=N=Nx$, and for $x\notin N$, $xN=G\setminus N=Nx$.